If a, b, c are in AP , then the value of the determinant | array lll x+2 & x+3 & x+2 a x+3 & x+4 & x+2 b x+4 & x+5 & x+2 c array | is

Step-by-step solution

Let = | array lll x+2 & x+3 & x+2 a x+3 & x+4 & x+2 b x+4 & x+5 & x+2 c array | Applying R₂ 2 R₂ , we get = 1 2 | array ccc x+2 & x+3 & x+2 a 2 x+6 & 2 x+8 & 2 x+4 b x+4 & x+5 & x+2 c array | Applying R₂ R₂- (R₁+R₃ ) , we get = 1 2 | array ccc x+2 & x+3 & x+2 a 0 & 0 & 4 b-(2 a+2 c) x+4 & x+5 & x+2 c array | = 1 2 | array ccc x+2 & x+3 & x+2 a 0 & 0 & 0 x+4 & x+5 & x+2 c array | [ array c a, b, c are in AP 2 b=a+c array ] =0